Answer:
79 is the ninth term.
Step-by-step explanation:
You add nine to each one.
7, 16, 25, 34, 43, 52, 61, 70, 79
79 is the ninth term.
Answer:
x=1+2/15
Step-by-step explanation:
4(5x-3)=5x+5=118
We move all terms to the left:
4(5x-3)-(5x+5)=0
We multiply parentheses
20x-(5x+5)-12=0
We get rid of parentheses
20x-5x-5-12=0
We add all the numbers together, and all the variables
15x-17=0
We move all terms containing x to the left, all other terms to the right
15x=17
x=17/15
x=1+2/15
At the time the rocket hits the ground h=0, given that h=-16t²+320t+32
when h=0, our equation will be:
-16t²+320t+32=0
solving the above by completing square method we proceed as follows;
-16t²+320t+32=0
divide though by -16 we get
t²-20t-2=0
t²-20t=2
but
c=(-b/2a)^2
c=(20/2)^2
c=100
hence:
t²-20t+100=100+2
(t-10)(t-10)=102
√(t-10)²=√102
t-10=√102
hence
t=10+/-√102
t~20.1 or -0.1
since it must have taken long, then the answer is 20.1 sec
Answer:
The absolute brightness of the Cepheid star after a period of 45 days is -5.95
Step-by-step explanation:
Since the absolute magnitude or brightness of a Cepheid star is related to its period or length of its pulse by
M = –2.78(log P) – 1.35 where M = absolute magnitude and P = period or length of pulse.
From our question, it is given that P = 45 days.
So, M = –2.78(log P) – 1.35
M = –2.78(log 45) – 1.35
M = –2.78(1.6532) – 1.35
M = -4.60 - 1.35
M = -5.95
So, the absolute magnitude or brightness M of a Cepheid star after a period P of 45 days is -5.95
